Since it's a projectile being launched the only force acting upon it is gravity, since the object is in free fall once it's launched
so to calculate time you'd utilize the general formula of

and then solve using time and make it into the y axis, so change the x's to y's, which will change a to g.
since Vyi is always usually 0, you can drop that out of this equation so the formula to find time would be

So you'll plug in and it'll be

to find the maximum height you'll have to do some trigonometry to solve it.
To make it easier draw a triangle
put the 60° mark as shown in the picture.
Then you'll need to find the hypotenuse or horizontal to find the vertical
So the hypotenuse would be the 113m/s
so then you'll use

plug in the numbers

now that you have the vertical
use the formula

solve for d which will give you the hypotenuse

The "vertical" is what you found in the previous step.
Vf^2 is equal to 0 so you can just drop that number out since it's 0
then once you have that then youre not done yet
since you're on a cliff of 49 m you'll have to add 49m to the previous answer that you found d to find the maximum height.
I hope this helps!
Answer:
a) 0, = -0.33 us
b) 140m
c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.
Explanation:
a)
the lorentz factor expression is written as;
y = 1₀ / √(1 - (v²/c²))
where v is the relative speed of an observer and c is the speed of light
so we were given that relative speed to be o.7c
therefore
y = 1 / √(1 - ((0.7c)² / c²))
y = 1 / √(1 - (0.49c² / c²))
y = 1 / √(1 - 0.49)
y = 1 / 0.7141
y = 1.4
1 - the coordinates of the first event, the s' frame of reference is,
x1 ' = y(x1 - vt1) = 0
y1 ' = y1, z1' = z1 and
t1 ' = y [t1 - v/c²x1]
= 0
2 - the coordinates of the second event, the s ' frame of reference is'
x2 ' = y(x2-vt2)
= 1.4(100m - 0)
= 140m
y2 ' = y2, z2 ' = z2
t2 ' = y [ t2 - v/c²x2 ]
= 1.4 [ 0 - 0.7c/c²(100) ]
using speed of light c as 3*10^8
1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]
= -0.33 us
b)
distance between
delltaX' = X2' - X1'
= 140m - 0
= 140m
c)
No, The event are not simultaneous i.e they did not occur at the same time.
the second even (-0.33 us) occurs 0.33 us earlier than the first event.
Answer:
No, they will not change.
Explanation: